If `x` is real and satisfies `x +2> sqrt(x +4)`, then

Options

A. `x lt - 2`
B. `x gt 0`
C. `-3lt x lt 0`
D. `-3 lt x lt 4`

Explanation

To solve the inequality \( x + 2 > \sqrt{x + 4} \), we will follow these steps:

### Step 1: Square both sides
We start with the given inequality:
\[ x + 2 > \sqrt{x + 4} \]
To eliminate the square root, we square both sides:
\[ (x + 2)^2 > x + 4 \]

### Step 2: Expand the left side
Now, we expand the left side:
\[ x^2 + 4x + 4 > x + 4 \]

### Step 3: Rearrange the inequality
Next, we rearrange the inequality to bring all terms to one side:
\[ x^2 + 4x + 4 - x - 4 > 0 \]
This simplifies to:
\[ x^2 + 3x > 0 \]

### Step 4: Factor the quadratic expression
Now, we factor the quadratic expression:
\[ x(x + 3) > 0 \]

### Step 5: Find the critical points
The critical points occur when the expression equals zero:
\[ x = 0 \quad \text{and} \quad x + 3 = 0 \Rightarrow x = -3 \]

### Step 6: Test intervals
We will test the intervals determined by the critical points: \( (-\infty, -3) \), \( (-3, 0) \), and \( (0, \infty) \).

1. **Interval \( (-\infty, -3) \)**: Choose \( x = -4 \)
\[ (-4)(-4 + 3) = (-4)(-1) = 4 > 0 \quad \text{(True)} \]

2. **Interval \( (-3, 0) \)**: Choose \( x = -1 \)
\[ (-1)(-1 + 3) = (-1)(2) = -2 < 0 \quad \text{(False)} \]

3. **Interval \( (0, \infty) \)**: Choose \( x = 1 \)
\[ (1)(1 + 3) = (1)(4) = 4 > 0 \quad \text{(True)} \]

### Step 7: Combine results
From our tests, the inequality \( x(x + 3) > 0 \) holds true in the intervals:
\[ (-\infty, -3) \quad \text{and} \quad (0, \infty) \]

### Step 8: Conclusion
Since \( x \) must be real, the solution to the inequality \( x + 2 > \sqrt{x + 4} \) is:
\[ x < -3 \quad \text{or} \quad x > 0 \]